Can a Force of Gravitation Be Considered Conservative?

[WiFO215]

I was watching 18.02 now and Denis Aurox, the prof. who is lecturing, mentioned that for a force to be conservative, not only was it necessary for its curl = 0 but also that F is defined and differentiable everywhere. He then went on to give the example y<i> + x<j>/x^2 + y ^2 will not be a conservative force for it is not defined at the origin.

However, I have a doubt. The first example that hit me was the Force of Gravitation. It's given by Const./ r^2 where r is the displacement between the masses. When we look at it in terms of a field, the field is not defined on the mass itself (i.e. on the point mass) and shoots to infinity as r shoots to zero. How come we define a potential in this case?

Anirudh

 

[Feldoh]

Curl is just a consequence of a conservative vector field. For some smooth function f which is defined on some domain D, then we can define a conservative vector such that F = grad(f).

So in your first example it is conservative, you just have to give a proper domain.

 

[HallsofIvy]

You are confusing "if" with "only if". If a vector field has curl 0 and is defined and differentiable everywhere, it is conservative. If a vector field does not satisfy those conditions, it may or may not be conservative.

 

[WiFO215]

You are confusing "if" with "only if". If a vector field has curl 0 and is defined and differentiable everywhere, it is conservative. If a vector field does not satisfy those conditions, it may or may not be conservative.

I see. So how do you verify if it is or isn't conservative? By finding out if it can be expressed as a gradient of a function?

So in the case of Gravitation, even though the field is not defined at r = 0, we get a gradient field V satisfying grad V = G field because ...?

 

[WiFO215]

Okay. I got it. I got it. I just went ahead a few lectures where he talks about this in more detail and that clarified things.

<<SO THIS IS DONE>>

Could someone finish up my other thread lurking below this one?